Answer:
9 and 10
Step-by-step explanation:
This problem calls for setting up a system of equations. Let's call the first number n and the second number s.
One number is 1 less than a second number (hint: 'is' typically means equals!):
[tex]n=s-1[/tex]
Twice a second number is 25 less than 5 times the first.
[tex]2s=5n-25[/tex]
Now that we have a system set up, we can use substitution to solve.
Substitute the 'n' variable in the second equation with the value that we gave it in the first equation.
[tex]2s=5(s-1)-25[/tex]
Now that we have only one variable, we can solve algebraically!
[tex]2s=5s-5-25\\2s=5s-30\\2s+30=5s\\30=3s\\s=10[/tex]
Now that we have the value of one variable, we can go back to the original equations and plug in s in order to solve for n.
[tex]n=s-1\\n=10-1\\n=9[/tex]
And there we have both numbers!
I hope this helps!
